Prove that 1square + 2square + 3square .......+nsquare =n(n+1)2n+1 divided by 6
Answers
Step-by-step explanation:
To prove ---->
--------------
n(n+1)(2n+1)
1²+2²+..................+n²=-----------------------
6
proof---> we prove it by pmi
---------
first we prove it for n=1
first term=1²=1
now putting n=1 in RHS
1(1+1)(2×1+1)
=---------------------
6
1(2)(2+1)
=--------------------
6
6
=--------=1
6
so given statement is true for n=1
now let given statement is true for n=k
k(k+1)(2k+1)
1²+2²+3²+........+k²=-----------------------
6
now we prove it is true for n=k+1
for this adding (k+1)² to both sides
k(k+1)(2k+1)
1²+2²+....+k²+(k+1)²=-------------------- +(k+1)²
6
k(2k+1)
=(k+1){--------------------------- +(k+1) }
6
k(2k+1) + 6 (k+1)
=(k+1) {_____________________}
6
2k²+ k + 6k + 6
= (k+1) (--------------------------------)
6
2k²+7k+6
= (k+1) (------------------------)
6
2k² +(4+3)k +6
=(k+1) {------------------------------}
6
2k²+4k+3k+6
=(k+1) {------------------------------}
6
2k(k+2) +3(k+2)
=(k+1) { _________________}
6
(k+2) (2k+3)
=(k+1) {________________}
6
{(k+1)+1}{(2k+2)+1}
=(k+1) [_____________________]
6
(k+1) {(k+1) +1 } {2(k+1)+1}
=--------------------------------------------------
6
so given statement is true for n=k+1 if it's true for n=k
so by principle of mathematical induction given statement is true for any natural number
Hope it helps you
Thanks for giving me chance to answer your question
another method is in attachment
Answer:
Step-by-step explanation:
We have that:
1² + 2² + 3² + ... + n² = n(n + 1)(2n + 1)/6
We can prove this by induction. First, we need to show that it is true for n = 1.
1(1 + 1)(2 + 1)/6 = 1
So it is true for n = 1.
Now, we need to show that if it is true for a value, then it is true for the next value (which makes it true for all integers greater than that value as well).
Adding (n + 1)² to both sides gives us:
1² + 2² + 3² + ... + n² + (n + 1)² = n(n + 1)(2n + 1)/6 + (n + 1)²
Focusing on the RHS:
n(n + 1)(2n + 1)/6 + (n + 1)²
= n(n + 1)(2n + 1)/6 + 6(n + 1)²/6
= [n(n + 1)(2n + 1) + 6(n + 1)²]/6
= (n + 1)[n(2n + 1) + 6(n + 1)]/6. . . . . . . . . .Factor out (n + 1)
= (n + 1)(2n² + n + 6n + 6)/6
= (n + 1)(2n² + 4n + 3n + 6)/6
= (n + 1)[2n(n + 2) + 3(n + 2)]/6
= (n + 1)(2n + 3)(n + 2)/6
= (n + 1)[2(n + 1) + 1][(n + 1) + 1]/6
Since this result is the same as before except n + 1 is in n's spot. It is true for all n equal to or greater than 1! Q.E.D.
I hope this helps!